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Electrostatics question

2023 · 12 Apr · Shift 1 · Q61

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2023 · 12 Apr · Shift 1 · Q61

Electrostatics question

2023 · 12 Apr · Shift 1 · Q61

JEE MainPhysicsElectrostaticsNumerical+4 / −1

64 identical drops each charged upto potential of

10 ~\mathrm{mV}

are combined to form a bigger drop. The potential of the bigger drop will be

\underline{\hspace{2cm}}

\mathrm{mV}

.

Numerical answer

View written solutionFree

Correct answer: 160

Potential of a charged spherical drop

For an isolated conducting spherical drop, $V = \frac{1}{4\pi\varepsilon_0}\frac{q}{r}$ So, potential is proportional to $V \propto \frac{q}{r}$

When 64 identical drops combine

Let each छोटे drop have:

radius $r$
charge $q$
potential $V = 10\,\text{mV}$

If $n=64$ identical drops combine:

Total charge of bigger drop: $Q = 64q$
Volume is conserved: $\frac{4}{3}\pi R^3 = 64\cdot \frac{4}{3}\pi r^3$ $R^3 = 64r^3$ $R = 4r$

Potential of the bigger drop

The new potential is

= \frac{1}{4\pi\varepsilon_0}\frac{64q}{4r}$$ $$V' = 16\left(\frac{1}{4\pi\varepsilon_0}\frac{q}{r}\right)$$ $$V' = 16V$$ Since $V=10\,\text{mV}$, $$V' = 16\times 10 = 160\,\text{mV}$$ 4. **Final answer** The potential of the bigger drop is $$\boxed{160\,\text{mV}}$$

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